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                <h1 class="title-article" id="articleContentId">Python实现一个简单的Feistel结构的密码算法</h1>
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                <div class="up-time"><span>最后发布:2016-03-27 15:29:59</span><span>首次发布:2016-03-27 15:29:59</span></div>
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                    <p>这是我们密码学老师留的一个作业&#xff0c;她本来留的作业内容是仿照DES密码设计一个密码算法&#xff0c;但第一次听到这个作业时我的内心是崩溃的&#xff08;冏<span style="font-family:Calibri">rz</span><span style="font-family:宋体">&#xff09;&#xff0c;因为其实我并不太能理解</span><span style="font-family:Calibri">DES</span><span style="font-family:宋体">算法中</span><span style="font-family:Calibri">P</span><span style="font-family:宋体">盒还有</span><span style="font-family:Calibri">S</span><span style="font-family:宋体">盒设计的思想是什么。。&#xff08;也就是并不太能理解这些置换表为什么要这么设计&#xff09;</span></p> 
<p><span style="font-family:宋体"></span></p> 
<p>但是我知道<span style="font-family:Calibri">DES</span><span style="font-family:宋体">算法是一种分组密码&#xff0c;采用的是</span><span style="font-family:Calibri">Feistel</span><span style="font-family:宋体">结构的呀&#xff01;于是我就根据</span><span style="font-family:Calibri">Feistel</span><span style="font-family:宋体">结构的设计思想设计了一个简单的加密算法。</span></p> 
<p>Feistel<span style="font-family:宋体">结构最大的优点就是加密和解密过程可以使用完全相同的算法&#xff0c;这就使得在实施的过程中&#xff0c;对编码量和线路传输的要求就减少了几乎一半。使用其结构的安全性主要在于选择块的大小&#xff0c;密钥长度&#xff0c;循环次数&#xff0c;子密钥生成算法和轮函数复杂度。</span></p> 
<p> </p> 
<p>我设计的算法中&#xff0c;把字符串的<span style="font-family:Calibri">8</span><span style="font-family:宋体">个字符分为一组加密&#xff0c;每次加密时使用</span><span style="font-family:Calibri">4</span><span style="font-family:宋体">个字符的字符串作为密钥&#xff08;一个</span><span style="font-family:Calibri">ASCII</span><span style="font-family:宋体">字符</span><span style="font-family:Calibri">8</span><span style="font-family:宋体">个</span><span style="font-family:Calibri">bit</span><span style="font-family:宋体">&#xff0c;所以</span><span style="font-family:Calibri">4</span><span style="font-family:宋体">个字符应该就是</span><span style="font-family:Calibri">32</span><span style="font-family:宋体">个</span><span style="font-family:Calibri">bit</span><span style="font-family:宋体">了吧⊙▽⊙&#xff09;&#xff0c;加密算法很简单&#xff0c;就是把每个分组</span><span style="font-family:Calibri">8</span><span style="font-family:宋体">个字符分为两组&#xff0c;每组的字符按位与密钥字符异或。</span></p> 
<p><span style="font-family:宋体"></span></p> 
<p>8<span style="font-family:宋体">个字符为一个分组&#xff0c;前</span><span style="font-family:Calibri">4</span><span style="font-family:宋体">个字符为设为</span><span style="font-family:Calibri">L0</span><span style="font-family:宋体">&#xff0c;后</span><span style="font-family:Calibri">4</span><span style="font-family:宋体">个字符设为</span><span style="font-family:Calibri">R0</span></p> 
<p>即设<span style="font-family:Calibri">L0&#61;</span>’aaaa’,R0&#61;’bbbb’,key&#61;’abcd’</p> 
<p>加密算法为</p> 
<p>For i in (1 to 4)</p> 
<p>L1[i]&#61;L0[i] xor R0[i] xor key[i]</p> 
<p>R1&#61;L0</p> 
<p> </p> 
<p>然后我用<span style="font-family:Calibri">Python</span><span style="font-family:宋体">实现了上述加密&#xff0c;但是程序有一个小瑕疵&#xff0c;处理的明文字符数必须为</span><span style="font-family:Calibri">8</span><span style="font-family:宋体">的整数倍&#xff0c;否则多出来的字符将不会处理。但这也无伤大雅啦&#xff0c;只是一些细节的处理而已。</span></p> 
<br /> 
<p><span style="font-family:宋体">可能我之前描述设计思想时讲地并不是很好&#xff0c;所以我在这里提供程序的完整代码&#xff0c;毕竟Linux之父Linus大神也说过“Talk is cheap. Show me the code”<br /> </span></p> 
<p><span style="font-family:宋体"></span></p>
<pre><code class="language-python"># -*- coding: utf-8 -*-
#每个分组8个字符&#xff0c;一个密钥4个字符

def strEncrypt(s1,s2,key):#每次传输8个字符,分成两组按位异或,密钥4个字符
    result&#61;&#39;&#39;
    for i in range(0,4):
        temp&#61;ord(s1[i])^ord(s2[i])^ord(key[i])
        result&#43;&#61;chr(temp)
    return result

def encode(plainText,keylist):
    cipherText&#61;&#39;&#39;
    for i in range(0,len(plainText)//8):
        temp&#61;plainText[i*8:(i&#43;1)*8]#每8个字符划分成一组加密
        L&#61;temp[:4]
        R&#61;temp[4:]
        for currentKey in keylist:
            temp2&#61;R
            R&#61;strEncrypt(L,R,currentKey)
            L&#61;temp2
        cipherText&#43;&#61;R&#43;L
    return cipherText

def main():
    key&#61;[]
    a&#61;input(&#39;请输入要加密的字符串&#xff0c;目前算法能处理的明文字符为8的整数倍&#39;)
    keynum&#61;input(&#39;请输入加密循环次数&#39;)
    for i in range(0,int(keynum)):
        temp&#61;input(&#39;请输入第&#39;&#43;str(i)&#43;&#39;组加密时使用的密钥&#xff08;4个字符&#xff09;&#39;)
        key.append(temp)
    print(&#39;要加密的明文为&#xff1a;&#39;,a)
    a&#61;encode(a,key)
    print(&#39;密文&#xff1a;&#39;,a)

    #####解密
    key.reverse()
    a&#61;encode(a,key)
    print(&#39;解密后的明文&#xff1a;&#39;,a)


if __name__ &#61;&#61; &#39;__main__&#39;:
    main()</code></pre>
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给明文分组(L,R)

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很好理解对吧=_+，分组加密交换即可，我们再将其复杂化：

完整的<em>Feistel</em><em>密码</em>


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                    <a href="https://blog.csdn.net/weixin_50561472" class="user-name" target="_blank">马斌若: </a>
                    <span class="code-comments">想问一下，计算机不会先将加出来的结果存起来，再进行减法操作吗？溢出的话结果就可能不一样了。我这个想法是不是因为没理解这个阿贝尔群或者什么的啊，自己看的不是太懂，有时间可以解释一下吗</span>
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                    <a href="https://blog.csdn.net/qq_50912555" class="user-name" target="_blank">qq_50912555: </a>
                    <span class="code-comments">可以链接锁，锁的灯变蓝了打不开</span>
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                <a class="title text-truncate" target="_blank" href="https://blog.csdn.net/CarlCurist/article/details/85345941#comments_13242692" data-report-click='{"mod":"popu_542","spm":"1001.2101.3001.4231","dest":"https://blog.csdn.net/CarlCurist/article/details/85345941#comments_13242692","ab":"new"}' data-report-view='{"mod":"popu_542","spm":"1001.2101.3001.4231","dest":"https://blog.csdn.net/CarlCurist/article/details/85345941#comments_13242692","ab":"new"}'>QIUI囚爱男用APP远程贞操锁2.0 破解不完全指南（附破解工具）</a>
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                    <span class="code-comments">求对新款破解一下</span>
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                    <a href="https://blog.csdn.net/weixin_50176623" class="user-name" target="_blank">戈彦华牵挂你: </a>
                    <span class="code-comments">我的锁是自己管</span>
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